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<h3 class="heading"><span class="type">Paragraph</span></h3>
<ol class="decimal">
<li>
<p>If <span class="process-math">\(\lambda=-k^2&lt;0\text{,}\)</span> then</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
y(x)=c_1e^{kx} + c_2e^{-kx},\quad y'(x)=kc_1e^{kx} - kc_2e^{-kx}
\end{equation*}
</div>
<p class="continuation">To determine <span class="process-math">\(c_1\text{,}\)</span> <span class="process-math">\(c_2\text{,}\)</span> use the boundary conditions:</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
kc_1-kc_2=0,\quad kc_1e^{kL}-kc_2e^{-kL}=0,\quad\Rightarrow\quad c_1=c_2=0
\end{equation*}
</div>
<p class="continuation">which gives only trivial solution (Discard).</p>
</li>
<li>
<p>If</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
y(x)=Ax+B,\quad\to\quad y'(x)=A
\end{equation*}
</div>
<p class="continuation">By boundary conditions, we have <span class="process-math">\(A=0,~B\in\mathbb{R}\text{.}\)</span> So we found an eigenpair:</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
{\lambda_0=0,\qquad y_0(x)=1}.
\end{equation*}
</div>
</li>
<li>
<p>If <span class="process-math">\(\lambda=k^2&gt;0\text{,}\)</span> then <span class="process-math">\(r^2 =-k^2~\to~r=\pm ki\text{,}\)</span></p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
y(x)=c_1\cos {kx} + c_2\sin {kx},\quad y'(x)=-kc_1\sin kx + kc_2\cos kx.
\end{equation*}
</div>
<p class="continuation">We now check the boundary conditions:</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
\left\{\begin{array}{l} y'(0)=kc_2=0\\
y'(L)=-kc_1\sin kL=0\end{array}\right.\Rightarrow \left\{\begin{array}{l} c_2=0\\
c_1\sin kL=0\end{array}\right.
\end{equation*}
</div>
<p class="continuation">If <span class="process-math">\(c_1 = 0\text{,}\)</span> then <span class="process-math">\(y(x) \equiv 0\)</span> which is trivial. So <span class="process-math">\(c_1\neq 0\)</span> and we must have <span class="process-math">\(\sin kL=0\text{:}\)</span></p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
kL=n\pi,\quad \Rightarrow \quad k=\frac{n\pi}{L},~~n=1,2,3,\cdots
\end{equation*}
</div>
<p class="continuation">For each <span class="process-math">\(k\text{,}\)</span> we get a pair of eigenvalue and eigenfunction</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
{\lambda_n}={\left(\frac{n\pi}{L}\right)^2},\quad {y_n(x)}={\cos\frac{n\pi x}{L}},\quad n={1},2,3,\cdots
\end{equation*}
</div>
</li>
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<span class="incontext"><a href="sec7_2.html#p-326" class="internal">in-context</a></span>
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